\section{Choosing the endpoints}\label{sec:endpoints}

\subsection{A uniform choice for every count}

For fixed $B,r,u$ and $e_*$, the numerator contribution to
$\tfrac12\log U$ in \eqref{eq:U} is proportional to $m$.
Differentiation adds the denominator cost $(m-1)\log e_*$.
Put $a_B=C_B-1$. We take the pole order $a=ma_B$ to offset
both costs. Its endpoint is
$a+m-1=m(a_B+1)-1=C_Bm-1$, linear in $m$.
The following estimates show that this scale suffices.

Take $r=2$, $L=5$, $u=1/100$, $e_0=68/25$ and $a=ma_B$.
Every denominator $d$ is odd, so its prime divisors are at
least $3$ and $\lceil5/(p-1)\rceil\le3$. Consequently
\begin{equation}\label{eq:K-bound}
 A_d=(2d)^5\prod_{p\mid d}p^{\lceil5/(p-1)\rceil}
       \le2^5d^8,\qquad
 K\le2^{5D}M^{8D}.
\end{equation}
We have $a_B\ge18D>5D$. The exact rational inequality
\begin{equation}\label{eq:f18}
 1001\left(\frac{401}{601}\right)^{18}<1
\end{equation}
therefore gives
\[
 f(u)\le
 \left[1001\left(\frac{401}{601}\right)^{18}\right]^{mD}<1.
\]

Put $A=\log(27/(4e_0))$. The exponential series gives
$163/60<\ep<68/25$. The following exact comparisons control
the numerator growth and the decay gained from $a$:
\begin{align}
 \frac{1001}{50}&<\left(\frac{163}{60}\right)^3,
       &\log(20.02)&<3,\label{eq:log20}\\
 4^{10}e_0^{19}&<27^{10},
       &A&>\frac9{10},\label{eq:slope}\\
 e_0&<\sum_{j=0}^5\frac{(11/10)^j}{j!},
       &\log e_0&<\frac{11}{10}.\label{eq:log-e0}
\end{align}
For \eqref{eq:slope}, the rational comparison implies
$10A>9\log e_0>9$.

Using \eqref{eq:U} and \eqref{eq:K-bound},
\begin{align*}
 \frac12\log U
 &=m\bigl(\log K+5D\log(10.01)-Aa_B\bigr)
                                      +(m-1)\log e_0\\
 &<m\left(15D+8D\log M-\frac9{10}a_B\right)
                                      +\frac{11}{10}(m-1).
\end{align*}
Definition~\eqref{eq:closed-endpoint} gives
$a_B\ge18D+(80/9)D\log M$.
The coefficient $80/9$ cancels the logarithmic cost because
$(9/10)(80/9)=8$. The constant $18$ leaves a margin of
$(6/5)mD$, since $(9/10)18-15=6/5$. Because $D\ge1$, this
absorbs the derivative cost $(11/10)(m-1)$. Indeed,
\[
 \frac12\log U
 <-\frac65mD+\frac{11}{10}(m-1)
 \le-\frac m{10}-\frac{11}{10}<0.
\]
These choices give the coefficients $9$ and $40/9$ in
\eqref{eq:closed-endpoint}; the factor $2$ ensures that
$a_B=C_B-1$ is even.
Proposition~\ref{prop:criterion} proves Theorem~\ref{thm:intervals}.
At $B=1$, we have $D_1=1$, $a_1=18$ and $C_1=19$, giving
Theorem~\ref{thm:one}.

\subsection{Sharper individual intervals}\label{sec:finite-endpoints}

For the individual intervals below, use the tighter rational upper bound
\begin{equation}\label{eq:e-tight}
 e_*=\sum_{j=0}^{20}\frac1{j!}+\frac1{20\cdot20!}>\ep.
\end{equation}
The tail bound follows by replacing all subsequent factorial
ratios by a geometric series.

The following choices satisfy $a>mDL$ and the two rational
inequalities in Proposition~\ref{prop:criterion}.
All rows use $r=2$, the bound \eqref{eq:e-tight}, and
$u=t/10^{12}$ with the integer $t$ shown.
The endpoint is $s=a+m-1$.
\begin{center}
\begin{tabular}{@{}rrrrrr@{}}
\toprule
$m$&$B$&$a$&$s$&$t$&$U$ (approx.)\\
\midrule
1&2&70&70&779643543&0.09965\\
1&3&186&186&209514585&0.03463\\
1&4&360&360&132973689&0.96654\\
1&5&554&554&73388877&0.16375\\
1&10&2794&2794&11811809&0.26283\\
3&1&52&54&8990232707&0.49983\\
5&1&88&92&8058760219&0.10513\\
7&1&122&128&8645692609&0.84652\\
19&1&334&352&8128616097&0.10743\\
19&2&1326&1344&802179083&0.58430\\
19&3&3520&3538&218654559&0.08821\\
99&1&1740&1838&8139253515&0.07226\\
\bottomrule
\end{tabular}
\end{center}
The decimals illustrate the margins; all comparisons are exact
over $\Q$. These rows prove the individual intervals in the
introduction. For example, $m=19$, $B=2$ gives two irrational
values among $\beta(20),\beta(22),\ldots,\beta(1344)$.
The accompanying endpoint file contains $21$ certificates in
total, including additional choices for counts $4$, $5$ and $10$.